Concecutives square

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x ( x + 1 ) ( x + 2 ) ( x + 3 ) + 1 x(x+1)(x+2)(x+3)+1 where x x is an integer is a perfect square of the form ( a x 2 + b x + c ) 2 (ax^2+bx+c)^2 .Let f ( x ) = a x 2 + b x + c f(x)=ax^2+bx+c .Find f ( 1 ) f(1) .


The answer is 5.

Lorenc Bushi by Lorenc Bushi , 21, Albania

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1 solution

Pi Han Goh
Dec 24, 2013

Note that product of four consecutive numbers is one less than a perfect square

x ( x + 1 ) ( x + 2 ) ( x + 3 ) + 1 = 1 + x ( x + 3 ) ⋅ ( x + 1 ) ( x + 2 ) = 1 + ( x 2 + 3 x ) ( x 2 + 3 x + 2 ) = 1 + ( x 2 + 3 x + 1 − 1 ) ( x 2 + 3 x + 1 + 1 ) = 1 + ( x 2 + 3 x + 1 ) 2 − 1 2 , by difference of squares = ( x 2 + 3 x + 1 ) 2 \begin{aligned} x(x+1)(x+2)(x+3) + 1 & = & 1 + x(x+3) \cdot (x+1)(x+2) \\ & = & 1 + (x^2 + 3x)(x^2 + 3x + 2) \\ & = & 1 + (x^2 + 3x + 1 - 1)(x^2 + 3x + 1 + 1) \\ & = & 1 + (x^2 + 3x + 1)^2 - 1^2, \text{by difference of squares} \\ & = & (x^2 + 3x + 1)^2 \\ \end{aligned}

⇒ f ( x ) = x 2 + 3 x + 1 ⇒ f ( 1 ) = 1 2 + 3 ( 1 ) + 1 = 5 \Rightarrow f(x) = x^2 + 3x + 1 \Rightarrow f(1) = 1^2 + 3(1) + 1 = \boxed{5}

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